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Mathematics (2015) QUIZ STUDY
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1.
In a town of 6250 inhabitants, there were 62 births during 1984. Find the percentage birth rate
.
A. 3%
B. 1.00%
C. 2.50%
D. 5.40%
Explanation.
Percentage birthrate = 1.0%

2.

Simplify 1¼ ÷ (2 ÷ ¼) of 28

.
A.

1 \( \frac{3}{8}\)

B.

2 \( \frac{3}{4}\)

C.

4 \( \frac{3}{8}\)

D.

3 \( \frac{1}{5}\)

Explanation.

1¼ ÷ [ 2 ÷ ¼] of 28

Apply BODMAS rules

\( \frac{5}{4}\) ÷ [ 2 ÷ ¼ \(\times\) 28 ]

\( \frac{5}{4} ÷ \frac{2}{7} \)


\( \frac{5}{4} \times \frac{7}{2} \)

\( \frac{35}{8}\)

= \(4 \frac{3}{8}\)


3.

Factorize x2 + 9x + 20

.
A.

(x - 5) 2

B.

(x + 5)(x + 4)

C.

(x + 5)(x + 3)

D.

(x + 3) 2

Explanation.

\( x^2 + 9x + 20 \)

Find the two numbers whose product is 20 and its sum is 9}

\( 5x \times 4x = 20x^2\)


\(5x + 4x = 9x\)

\((x^2 + 5x) + (4x + 20)\)
\(

x(x + 5)+ 4(x + 5)

= (x + 5)(x + 4)\)


4.

If three staff of QuizzerWeb Inc. agreed to share their salary arrears in the ration of their ages, which are 18 years, 20 years, 22 years respectively. If the sum of the money collected is N120,000.00K, How much does the second staff received?

.
A.

N36,000

B.

N44,000

C.

N40,000

D.

N15,000


5.

X and Y are two sets such that nX = 15, nY = 12 and n{- Y} = 7. Find -{X - Y}

.
A.

21

B.

225

C.

15

D.

20

Explanation.

n(X - Y) = nx + ny - n(xX - Y)

= 15 + 12 - 7

- n(X - Y) = 20


6.

Find the x and y intercepts of the graph of 3x - z \(\leq\) 9

.
A.

(3, -9)

B.

(-3, 9)

C.

(-3, -9)

D.

(-3, 0)

Explanation.

Starting from 3x - z \(\leq\) 9

x = 0, substitute the value of x (i.e. x = 0) into the equation 3x - z \(\leq\) 9

3(0) - z \(\leq\) - 9

z \(\leq\) - 9

If z = 0

Then, 3x - 0 \(\leq\) 9

3x \(\leq\) 9

x \(\leq\) 3

Intercept of x and z i.e (x,z)

= (3, -9)


7.

Simplify log101.5 + 3 log102 - log100.3

.
A.

log104

B.

log1040

C.

log10-40

D.

log104-

Explanation.

log101.5 + 3 log102 - log100.3

log101.5 + log1023 - log100.3

[log10(1.5 \(\times\) 23) ÷ log100.3]

[log10(15/10 \(\times\) 8 ) ÷ log10( 3/10)]

log10(15/10 \(\times\) 8 \(\times\) 10/3 )

log1040


8.

Find the total surface area of a cylinder of base radius 5cm and length 7cm ( π = 3.14)

.
A.

17.8cm2

B.

15.8cm2

C.

75.4cm2

D.

54.7cm2

Explanation.

The total surface area of a cylinder = 2πrl + 2πr2

= 2πr(l + r)

= 2 \(\times\) 3.14(7+%=5)

2 \(\times\) 3.14 \(\times\) 12

= 75.4cm (1DP)


9.

If x + y = 90 simplify (sinx + siny)2 - 2sinxsiny

.
A.

1

B.

\( 0 \)

C.

2

D.

-1


10.

A man with an annual salary of N2000, has allowances of N600. If Income Tax is 5%. How much tax does he pay each year?

.
A.

15

B.

20

C.

30

D.

25

Explanation.

His annual salary = N2000

His allowances = N600

So his taxable income = Annual salary - allowance

= N2000 - N600

= N1400

He pay at 5%

Then, his allowance income tax 5/100 \(\times\) 600 = N30


11.

Solve the equation \( 3x^2 - 4x - 5 = 0 \)

.
A.

x = 1.75 or - 0.15

B.

x = 2.12 or - 0.79

C.

x = 1.5 or - 0.34

D.

x = 2.35 or -1.23


12.

The first and last term of a linear sequence (AP) are 6 and 10 respectively. If the sum of the sequence is 40. Find the number of terms

.
A.

nth = 3

B.

nth = 4

C.

nth = 5

D.

nth = 6

Explanation.

Nth term of a linear sequence (AP) = a+(n - 1)d

first term = 6, last term = 10 sum - 40

i.e. a = 6, l = 10, S = 40

sn = n ÷ 2(2a + (n - 1)d or Sn = ÷2 (a + b)

Sn = n ÷ 2(a + l)

40 = n ÷ 2(6 + 10)

40 = 8n

8n = 40

8n = 40

N = 40/8

= 5

The nth term = 5


13.

Find the equation of a line which is form origin and passes through the point (-3, -4)

.
A.

y = \( \frac{3x}{4} \)

B.

y = \( \frac{4x}{3} \)

C.

y = \( \frac{2x}{3} \)

D.

y = \( \frac{x}{2} \)


14.
The amount A to which a principal P amounts at r% compound interest for n years is given by the formula A = P(1 + (r ÷ 100))n. Find A, if P = 126, r = 4 and n = 2.
.
A. N132.50K
B. N136.30K
C. N125.40K
D. N257.42K
Explanation.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)


=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k

15.

Make x the subject of the equation
s = 2 + \(\frac{t}{5} \)(x + -y)

.
A.

x = 5[(s - 2) ÷ t] + 3yt

B.

x = 25[(s - 2) ÷ t] - 3ty

C.

x = [1 ÷ (s - 2)3ty]

D.

x = [5(s - 2) 2 ÷ 3ty] \(\times\) t

Explanation.

2 + t/5(x + 3/5y ) = s

t/5(x + 3/5y ) = s - 2

[(t(5x + 3y) ÷ 25)] = s - 2

t(5x + 3y) = 25 (s - 2)

5x + 3y = [(25(s - 2))÷ t]

5x = [(25(s - 2))÷ t] - 3y

Divide both sides by 5

x = [25(s - 2) - 3ty)]÷ 5t


16.
If log520 = x, find x
.
A. 1.761
B. 1.354
C. 1.861
D. 2.549
Explanation.
log520 = x

5x = 20(Take log10 of both sides)

log5x = log20

xlog5 = log20

x= [log20 ÷ log5]

[1.30103 ÷ 0.69897]

x = 1.861

17.

Find at which rate per annum simple interest N525 will amount to N588 in 3 years.

.
A.

3%

B.

2%

C.

5%

D.

4%

Explanation.

I = A - P

= N588 - N525

-I = N63

I = PRT ÷ 100

R = [100I ÷PT]

R = [(100 \(\times\) 63 ) ÷ (525 \(\times\) 3)]

= (6300 ÷ 1575) = 4

- The rate = 4 %


18.

Find the range of values of t which satisfies the inequality.

2t - 1 \(\leq\) 3 and 2 - t \(\geq\) 5

.
A.

-3 \(\leq\) t \(\leq\) 1

B.

-2 \(\leq\) t \(\leq\) 3

C.

-3 \(\leq\) t \(\leq\) 4

D.

-3 \(\leq\) t \(\leq\) 2

Explanation.

2t - 1 \(\leq\) 3 and 2 - t < 5

2t \(\leq\) 3 + 1 and 2 - 5 \(\leq\) t

2t \(\leq\) 4 and - 3 \(\leq\) t

t \(\leq\) 2 and - 3 \(\leq\) t or t \(\geq\) - 3

Combining the two ,starting from the last

-3 \(\leq\) t \(\leq\) 2


19.

Given that A = {1, 5, 7}

B = {3, 9, 12, 15}

C = {2, 4, 6, 8}

Find (A \(\cup\) B) \(\cup\) C

.
A.

{1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

B.

{1, 2, 3, 5, 6, 8, 12, 15}

C.

{2, 4, 5, 9, 12, 15}

D.

{1, 5, 6, 7, 8, 9, 12, 15}

Explanation.

{A \(\cup\) B \(\cup\) C}

{A \(\cup\) B} = {1, 3, 5, 7, 9, 12, 15}

 C={2,4,6,8}

{A \(\cup\) B} \(\cup\) C = {1, 3, 5, 7, 9, 12, 15} \(\cup\) {2, 4, 6, 8}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

{A \(\cup\) B} \(\cup\) C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}


20.

The extension of a stretched string is directly proportional to its tension. If the extension produced by a tension of 8 Newton's is 2cm, find the extension produced by a tension of 12 newton's.

.
A.

2

B.

1

C.

\( 0\)

D.

3


21.

Factorize \( x^2 - 2x - 15 \)

.
A.

(x + 3)2

B.

(x + 5)(x - 3)

C.

(x - 5)2

D.

(x - 5)(x + 3)

Explanation.

x2 - 2x - 15

(x2 - 5x) + (3x - 15)

x(x - 5)+ 3(x - 5)

(x - 5)(x + 3)


22.

If A = (-3,5) and B = (4,-1) find the co-ordinate of the mid point

.
A.

2, ½

B.

½ , 2

C.

1, ½

D.

0, 2


23.
Find the simple interest on N325 in 5years at 3% per annum.
.
A. N48.75K
B. N50.10K
C. N75.50K
D. N15.75K
Explanation.
I =[ PRT ÷ 100]

[(325 \(\times\) 5 \(\times\) 3) ÷ (100)]

N(195 ÷ 4)

N48.75K

24.
Simplify (0.09)2 and give your answer correct to 4 significant figures
.
A. 0.81
B. 0.081
C. 0.0081
D. 8.0001
Explanation.
(0.09)2 = 0.09 \(\times\) 0.09

= 0.0081

= 0.008100 to 4 significant figures

Please, start counting first from the non-zero digits i.e. 8

25.
Simplify \( [1 ÷ (x^2 + 3x + 2)] + [1 ÷ (x^2 + 5x + 6)] \)
.
A. \( \frac{2}{(x + 1)^2} \)
B. \( \frac{2}{(x + 1)(x + 2} \)
C. \( \frac{2}{(x + 1)(x + 2} \)
D. \( \frac{2}{(x + 1)(x + 3} \)
Explanation.
\( [1 ÷ (x^2 + 3x + 2)] + [1 ÷ (x^2 + 5x + 6)] \)

= \( 1 ÷ (x^2 + 3x + 2) + [1 ÷ (x^2 +5x + 6)]\)

= \( [1 ÷ ((x^2 + x) + (2x + 2) )] + [1 ÷ ((x^2 + 3x) + (2x + 6) )] \)

= [1 ÷ (x(x + 2) + 2(x +1))] + [1 ÷ (x(x + 3) +2(x + 3) )]

= [1 ÷ (x + 1)(x + 2)] + [1 ÷ ((x + 3) + (x + 2))]

=((x + 3) + (x + 1)) ÷ (x + 1)(x + 2)(x + 3)

Using the L.C.M

=((x + x + 3 + 1)) ÷ (x + 1)(x + 2)(x + 3)

=(2x+4)/(x+1)(x+2)(x+3) =2(x+2)/(x+1)(x+2)(x+3)

= \( \frac{2 }{(x + 1)(x + 3)} \)

26.

One bag contain 3 blue and 5 red balls, another bag contain 2 blue and 4 red balls respectively. One ball is drawn for each bag. What is the probability both balls are blue

.
A.

\( \frac{2}{15} \)

B.

\( \frac{3}{24} \)

C.

\( \frac{3}{21} \)

D.

\( \frac{3}{28} \)

Explanation.

One bag with 3 blue and 5 red

Pr (H) = 3/8Pr (r) 5/8

Another bag with 2 blue and 4 red (note one ball is drawn from each bag)

i.e. prob.=8 - 1 = 7

prob. (2 ball both are blue=pr (1st blue and 2nd blue)

prob. 2 balls both blue = 3/8 \(\times\) 2/7

= \( \frac{3}{28} \)


27.

Given that A = {3, 4, 1, 10, ? }

B = {4, 3 ?, ?, 7}.

Find A ∩ B

.
A.

{}

B.

{?, 1, 2}

C.

{3, 4, ?}

D.

{1, 3, ?}

Explanation.

A B = {3, 4, ?}


28.

Find x if 132x = 70 eight.

.
A.

5

B.

3

C.

6

D.

1

Explanation.

132x = 708

\( 1 \times x^2 + 3 \times x^1 + 2 \times x^0 \)

\( 7 \times 8^1 + 0 \times 8^0 \)

\( x^2 + 3x + 2 - 56 = 0 \)

\( x^2 + 3x - 54 = 0 \)

x(x + 9)-6(x + 9) = 0

(x + 9)(x - 6) = 0

Either (x + 9) = 0 or (x - 6) = 0

x = - 9 or +6

The positive values for x = 6

The base number for 132x = 1326


29.

A man sells his new brand car for N420,000 at a gain of 15%. What did it cost him?

.
A.

N410,000

B.

N365, 217

C.

N157, 250

D.

N257,000

Explanation.

Cost Price:

Selling price = 100 : (100 + 15)

= 100 : 115

Cost Price

= 100 ÷ 115 of selling price (i.e N420,000)

= 100 ÷ 15 \(\times\) N420,000

= N365,217.4

Cost Price = N365, 217


30.

Find the value of x if \( [ 1 ÷ 64^{(x + 2)}]= [4^{(x ? 3)} ÷ 16^x ] \)

.
A.

\( \frac{3}{2} \)

B.

\( \frac{2}{3} \)

C.

\( \frac{1}{3} \)

D.

\( -\frac{3}{2} \)

Explanation.

\( [ 1 ÷ 64^{(x + 2)}]= [4^{(x - 3)} ÷ 16^x ] \)

\( 64^{-(x + 2)} = [4^{(x - 3)}] ÷[16^x] \)

breakdown 4,16,64 into a small index no

\( 2^{-6(x + 2)} = 2^{2(x - 3)} ÷ 2^{4(x)} \)

\( 2^{-6x- 12} = 2^{2x - 4x - 6} \)

\( 2^{-6x -12} = 2^{-2x - 6} \)

- 6x - 12 = - 2x - 6

Collect the like term

-6x + 2x = -6 + 12

-4x =6

x = \( \frac{6}{4} \)

x = \( \frac{-3}{2} \)


31.

A pipe made of metal 10cm thick has an external radius of 11cm. find the area of metal in making 2.4cm of pipe

.
A.

21πcm2

B.

21πcm2

C.

15πcm2

D.

17πcm2

Explanation.

The external radius = 11cm

The internal radius = 10cm

The area of cross section = π(π2 - 102)

= π (11 + 10)(11 - 10)

π(21)(1)

= 21πcm2


32.

Solve x2 - 2x - 3 = 0

.
A.

x = 2 or 1

B.

x = 3 or -1

C.

x = 1 or 0

D.

x = 1 or 2

Explanation.

x2 - 2x - 3 = 0

-3x + x = - 2x

- 3x \(\times\) x = - 3x2

(x2 - 3x)(x - 3) = 0

(x - 3)(x + 1) = 0

Either (x - 3) = 0 or (x + 1) = 0

x = 3 or - 1


33.

The volume of a cylinder whose height is 4cm and whose radius 5cm is equal to (π = 3.14)

.
A.

3.13cm2

B.

145cm2

C.

314cm2

D.

214cm2

Explanation.

€œExplanation

V = πr2h

V = ?, h = 4cm, r = 5cm

V = 3.14(5)2 (4)

V = 3.14(25)(4)

V = 3.14 \(\times\) 1200

= 314cm2


34.

\( \int x^2 \text{Sin}x \Delta x \)

.
A.

-Cos x + 2xSin x + C

B.

-Cos x + 2

C.

x2Cos x + 2xSin x + 2Cos x+ C

D.

-x2Cos x + 2xSinx + 2Cos x + C

Explanation.

Let u= x2

du = 2xdx

Let v = sinx

-dv = - sin x i.e. v = cos x

- udv = uv - - vdu

- x2 Sin xdx = - x2 Cos x - - (- Cos x)2 \(\times\) dx

x2 Cos x + 2 - x cos x1 dx

= - x2 Cos x + 2 - x Cos x2 dx

2 - u Cos xdx

Let u = x

du = dx

dv = Cos x

-dv = - Cos x

- x2 Sin x1 dx = - x2 Cos x + 2Sin x - 2Cos x + C

- x2 Sin xdx

= - x2 Cos x + 2xSin + 2Cos x + C


35.

The probability of an event A is 1/5. The probability of B is 1/3 . The probability both A and B is 1/15. What is the probability of either event A or B or both

.
A.

\( \frac{2}{15} \)

B.

\( \frac{3}{4} \)

C.

\( \frac{7}{15} \)

D.

\( \frac{1}{15} \)

Explanation.

Prob(A) = \( \frac{1}{5} \) , Prob(B) = \( \frac{1}{4} \), Prob (A  B) = \( \frac{1}{15} \), Prob (A B) = ?

Note:

I. the probability is either event of A or B or both.

The formula using is prob (A \(\cup\) B) = -prob(A) + prob(A) \(\cup\) prob(A B)

II. But if the probability of both outcomes A and B

The formula using is Prob(A B) \(\cup\) prob(A) + prob(B) + prob (A \(\cup\) B)

In this question, A or B, Prob (A \(\cup\) B): A and B, Prob (A \(\cup\) B)

prob (A \(\cup\) B) = prob(A) + prob(B) \(\cup\) Prob (A B) is used.

prob (A \(\cup\) B) = \( \frac{1}{15} \) + \( \frac{1}{3} \) + \( \frac{1}{15} \)

= (3 + 5 + 1) ÷ 15

= \( \frac{7}{15} \)


36.

Simplify 4 - [(390695T - 8)½)]

.
A.

1/T

B.

2T

C.

5/T

D.

T/5

Explanation.

4-(390625T - 8)½
[(390625T - 8)½]¼

[((390625)½ \(\times\) T - 8)½]¼

(- 390625 \(\times\) T - 4 )¼

(625 \(\times\) T -¼)¼

(54 \(\times\) T -4)¼

= (54)¼ \(\times\) (T - 4 )¼

= 5 \(\times\) T - 1

= 5 \(\times\) 1/T

= 5/T


37.
If (159.75)10 = (y)6. Find x
.
A. x6 = 123.346
B. x6 = 424.56
C. x6 = 122.436
D. x6 = 124.456
Explanation.
(159.75)10 = (y)6

\( \begin{array}{c|c}
6 & 159 \\
\hline
6 & 26 \text{ rem 3} \\
6 & 4 \text{ rem 2} \\
& 0 \text{ rem 4} \\
\end{array}\uparrow \)

15910 = 4236

Then Convert 0.75 to base 6

0.75 x 6 =4.5 i.e 4.5 - 3.00 = 1.5
0.5 x 6 = 3.00 423 + 1.5 =424.5

159.75 = (424.5)6

(159)10 = (424.5)6

38.

A man bought a car for N800 and sold it for N520. Find his loss per cent

.
A.

15%

B.

25%

C.

35%

D.

10%

Explanation.

Percentage Loss = (actual loss ) ÷ (Cost price) \(\times\) 100

Actual loss = Cost Price - Selling Price (sold price)

= N800 - N520 = N280

Percentage loss = (280 ÷ 800) \(\times\) 100 = 35%

⇒ His loss percent = 35%


39.

Simplify \(6\frac{1}{12} - 2 ¾ + 1 ½ \)

.
A.

\( 3 \frac{5}{6}\)

B.

\( 4 \frac{5}{6}\)

C.

\( 2 \frac{1}{6}\)

D.

\( \frac{1}{3} \)

Explanation.

6½ - 2¾ + 1½

73/12 - 11/4 + 3/2

[(73 - 33 + 18) ÷ (12)]

58/12

29/6

\( 4\frac{5}{6}\)


40.
Find the distance between the points (-2,-3) and (-2,4)
.
A. 3m
B. 2.4m
C. 3.2m
D. 5.1m

41.

Determine the third term of a geometrical progression whose first and second term are 2 and 14 respectively

.
A.

1458

B.

1485

C.

1345

D.

1258

Explanation.

1st G.P. = a =2

2nd G.P. = ar - 1 = 54

2(r) = 54

r = 54/2 = 27

r = 27

3rd term = ar2 = (2) (27)2

2 \(\times\) 27 \(\times\) 27

= 1458


42.

Given that S and T are sets of real numbers such that S = {x : 0 \(\leq\) x \(\leq\) 5} and T = {x : - 2 < x < 3} Find S \(\cup\) T

.
A.

-3 < x  \(\leq\)3

B.

-2 < x  \(\leq\) 5

C.

2< x \(\geq\) - 5

D.

-1 \(\leq\) x \(\geq\) 2

Explanation.

S = {0, 1, 2, 3, 4, 5}

T = {- 1, 0, 1, 2}

S \(\cup\) T = {- 1, 0, 1, 2, 3, 4, 5 }

- - 2 < x \(\leq\)  5


43.

If an investor invest N450,000 in a certain organization in order to yield X as a return of N25,000. Find the return on an investment of N700,000 by Y in the same organization.

.
A.

N14,950.50K

B.

N25,150.30K

C.

N15,000.00K

D.

N38,888.90K

Explanation.

[Return ÷ Investment] as a ratio ;

i.e The Ratio is Return : Investmen

[(Return1÷ Investment1 ) = (Return2÷ Investment2)]

R1 = N 25000

R2 =?

I1 = N450,000,

I1 = N 700000

(25000 ÷ 450000) = (R2 ÷ 700000)

R2 = [(25000 \(\times\) 700000 ) ÷ 450000]

= N38,888.90K

 The return on a investment of Y = N38888.90K


44.
Evaluate log717
.
A. 1.35
B. 1.353
C. 1.455
D. 0.455
Explanation.
log717

= [log 17 ÷ log7]

= [1.2304 ÷ 0.8451]

[100.0899 ÷ 101.9270]

= 1.455(antilog)

45.

100112 + *****2 + 111002 + 1012 = 10011112

.
A.

11112

B.

110112

C.

101112

D.

110012

Explanation.

Convert the binary to base 10 and they convert back to base two

100112 + xxxxx2 + 111002 + 1012 = 10011112


(1 \(\times\) 24 + 0 \(\times\) 23 + 1 \(\times\) 22 + 1 \(\times\) 21 + 1 \(\times\) 20) + xxxxx2 +(1 \(\times\) 24 + 1 \(\times\) 23 + 1 \(\times\) 22 + 0 \(\times\) 21 + 0 \(\times\) 20) + (1 \(\times\) 22 + 0 \(\times\) 21 + 1 \(\times\) 20)

= (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)

=(64 + 0 + 0 + 8 + 4 + 2 + 1)

19 + xxxxx2 + 33 = 79

xxxxx2 + 52 = 79

xxxxx2 = 79 - 52

xxxxx2 = 2710

\( \begin{array}{c|c}
2 & 27 \\
\hline
2 & 13 \text{ rem 1}\\
2 & 6 \text{ rem 1}\\
2 & 3 \text{ rem 0}\\
2 & 1 \text{ rem 1}\\
& 0 \text{ rem 1}\\
\end{array}\uparrow \)

2710 = 110112

Therefore xxxxx2 = 2710 = 110112


46.

Evaluate log5(\( y^2x^5 ÷ 125b½) \)

.
A.

2 log5y + 5log5 y2 - 3

B.

log5 y2 + 5log5 x + 3

C.

25logy 5 + 3

D.

2log5y + 5log5x - ½ log5b - 3


47.

If y = \((x^2 + 3x + 1) ÷ (3x + 4).\) Find dy/dx

.
A.

\( (3x^2 + 8x + 7) ÷ (3x + 4)^2 \)

B.

\( x^2 + 2x + (1 ÷ (x + 1)^2) \)

C.

\( 2x^2 + 4x + (1 ÷ (2x + 4)^2) \)

D.

\( x^2 + 3x + (2 ÷ (x + 3)^2) \)

Explanation.

dy/dx \( (4x^3 + 3x^2 + 2x + 1) \)

= \( 12x^2 + 6x + 2 \)

=  \( 12x^2 + 6 + 2 \)

Divide both sides by 2

\( \frac{12x^2}{2} + \frac{6x}2 + \frac{2}{2} \)

= \( 6x^2 + 3x + 1 \)

dy/dx

= \( 6x^2 + 3x + 1 \)


48.

Integral -\( (5x^3 + 7x^2 - 2x + 5)\)dx

.
A.

\( \frac{5x^4}{4} + \frac{7x^3}{3} + 2x + C \)

B.

\( \frac{5x}{4} + \frac{7x^3}{3} - x^2 + 5x + C \)

C.

\( \frac{5x^3}{3} + \frac{7x^2}{x} - x + C \)

D.

\( \frac{2x^2}{3} + \frac{x}{5} - C \)

Explanation.

\(- [(x^2 + 4x^2 + 1 ) ÷ x^2]dx = - (x^3/x^2)dx + -(1/x^2)dx \)

-\( - xdx + - 4^{-1} or + - (1/x^2)dx\)

= \( x^2/2 - 4x - 1/x^2 + C \)


49.

Find the area of the curved surface of a cone whose base radius is 3cm and whose height is 4cm (π = 3.14)

.
A.

17.1cm2

B.

27.2cm2

C.

47.1cm2

D.

37.3cm2

Explanation.

Find the slant height

\( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\)

\( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \)

\( l^2 =  25 \)

Squaring both sides

l = 5cm

The area of curved surface (s) =π(3)(5)

15π = 15 \(\times\) 3.14

= 47.1cm2


50.

Simplify \( \frac{1}{(x + 1)} + \frac{1}{(x - 1)} \)

.
A.

\(\frac{2x}{(x + 1)(x-3)} \)

B.

\(\frac{2}{(x + 1)(x-1)} \)

C.

\(\frac{2x}{(x + 1)2}\)

D.

2x(x+1)2

Explanation.

[1 ÷ (x+1)] + [1 ÷ (x - 1)]

= ((x - 1) + [(x + 1)) ÷ (x+1)(x - 1)]

Using the L.C.M.

= (x - 1 + x + 1) ÷ (x + 1)(x - 1)

= (x + 2 - 1 + 1) ÷ (x + 1)(x - 1)

= 2x ÷ (x + 1)(x - 1) =2x ÷ (x + 1)(x - 1)


51.

The area of a circle of radius 4cm is equal to (Take π = 3.142 )

.
A.

10.3cm2

B.

15.7cm2

C.

50.3cm2

D.

17.4cm2

Explanation.

Area of a circle (A) = πr2

Radius = 4cm

π = 3.142

A = 3.142 \(\times\) (4)2

= 3.142 \(\times\) 16

= 50.272

A = 50.3cm2 (1dp)


52.

The area of an ellipse is 132cm2.The length of its major axis is 14cm.Find the length of it minor axis

.
A.

10.5cm

B.

5cm

C.

10cm

D.

12cm

Explanation.

The area of an ellipse (e) =πab/4

Let a rep. the length of its major axis = 14cm

Let b rep. the length of its minor axis = ?

π = 3.142

Area of an ellipse = 132cm2

132 = (π14 \(\times\) b) ÷ 4

132 = (3.142 \(\times\) 14b) ÷ 4

3.142 \(\times\) 14b =132 \(\times\) 4

b = (1324 \(\times\) 4) ÷ (3.142 \(\times\) 14)
= 528/43.988

= 12cm

The length of its minor axis (b) = 12cm


53.

The volume of a cone (s) of height 6cm and base radius 5cm is

.
A.

157cm3

B.

155cm3

C.

175cm3

D.

145cm3

Explanation.

V = 1/πr2 h

Volume of a cone (Vc) = 1/ \(\times\) 3.14 \(\times\) (5)2 \(\times\) 6

= 156.9cm3

157cm3


54.

The probability of an outcome A is 1/6 . The probability of the B outcome is 1/4 . If the probability of A or B or both is 1/12 . What is the probability of both outcomes A and B?

.
A.

1/2

B.

1/3

C.

2/5

D.

3/4

Explanation.

Prob.(A outcome) = \(\frac{1}{6}\), Prob.(B outcome) = ¼

P(A B) = 1/12, Prob.(A \(\cup\) B)

Prob.(A \(\cup\) B) = Prob.(A) + Prob.(B) \(\cup\) Prob.(A B)

1/6 + 1/4 + 1/12

= (2 + 3 ? 1) ÷ 12

= 4/12

= 1/3

Prob.(A \(\cup\) B) = 1/3


55.
Given that Z = {1,2,4,5} what is the power of set Z?
.
A. 16
B. 8
C. 10
D. 12
Explanation.
Z has 4 elements, power = number of subset = 2p

Z = 2n = 24

= 16

56.

Calculate the value of x and y if 27x ÷ 81x+2y = 9 ,x + 4y = 0

.
A.

x = 1, y = 1/2

B.

x = 2, y =  1/2

C.

x = 0, y = 1

D.

x = 2, y = 1

Explanation.

\(27^x ÷ 81^{(x + 2y)} = 9 \)

\(27^x = 9 \times 81^{(x+2y)} \)

\({(3^{3} )}^x =32 \times 3^{4(x + 2y)} \)

\( =3^{(2 + 4x + 8y)}\)

\(3^{3x} = 3^{ (2 + 4x + 8y)}\)

3x = 2 + 4x + 8y

3x - 4x - 8y = 2 ----- (1)

x + 4y = 0 ------- (2)

- 4y = 2

y = (- 2) ÷ 4 = - ½

y = - ½

Substitute the value of y into equation (2)

i.e x + 4y = 0

x + 4( - 1/2) = 0

x - 2 = 0

x = 2

- x = 2,y = - ½

Method II

\( 27^x ÷ 31^{(x + 2y) }= 9\)

3^{3x} \(\times\) \(3^{( - 4x - 8y)} = 32\)

\( 3^{(3x - 8y)} = 32 \)

- x - 8y=2 ---- (1)

x + 4y = 0 ----- (2)

- 4 = 2

y= 2/4 = ½

y = ½ 

Substitute the value of y into equation 2

x + 4y=0

x + 4 (- 1) ÷ 2) = 0

x - 2 = 0

x = 2

x = 2, y = ½


57.

Simplify \(\sqrt{30} \times \sqrt{40} \)

.
A.

\(10 \sqrt{3}\)

B.

\(5 \sqrt{3}\)

C.

\(20 \sqrt{3}\)

D.

\(15 \sqrt{3}\)

Explanation.

\(\sqrt{30} \times \sqrt{40}\)

\(\sqrt{30} \times \sqrt{40}\)

\(\sqrt{3 \times 10 \times 4 \times 10}\)

\(\sqrt{400 \times 3}\)

\(\sqrt{400 \times 3}\)

\(20 \times \sqrt{3}\)

\(20 \sqrt{3}\)


58.

If ? (2x + 2) - -x =1 ,find x.

.
A.

x = 1 twice

B.

x = 0 or 1

C.

x = 3 or 1

D.

x = 2 twice

Explanation.

-(2x + 2 ) - -x = 1

-(2x + 2) = 1 + -x

Square both sides

-(2x + 2)2 = (1 + -x)2

((2x + 2)2)½ = 1 + -x + -x+ x

2x + 2 = 1 + 2-x +x

Collect the like term together

2x - x + 2 - 1 = 2 -x
x + 1 = 2-x

Square both sides

(x + 1)2 = (2-x)2

x2 + 2x + 1 = 4x

x2 + 2x - 4x + 1 = 0

(x2- x)(x + 1)= 0

x(x - 1) -1(x - 1)

(x - 1)(x -1)

Either (x - 1) = 0 or (x - 1) = 0

x = 1 or 1

x = 1 twice


59.
The sum of two numbers is 5; their product is 14. Find the numbers.
.
A. x = 5 or 1
B. x = 7 or 1
C. x = 7 or 0
D. x = 7 or ? 2
Explanation.
Let x represent the first number;

Then, the other is (5 ? x) , since their sum is 5 and their product is 14

x(5 ? x) = 14

5x ? x2 = 14

x2 ? 5x ? 14 = 0

(x2 ? 7x) + (2x ? 14) = 0

x(x ? 7) + 2(x ? 7) = 0

(x ? 7)(x + 2) = 0

Either (x ? 7) = 0 or (x + 2) = 0

x = 7 or x = ? 2

The two numbers are 7 and ? 2

60.

Solve the equation
\( 5^{(x - 2)} = (1 ÷ 125)^{(x +3)}\)

.
A.

3/-2

B.

-7/4

C.

4/6

D.

7/4

Explanation.

\( 5^{(x - 2)} = (125^{-1} )^{(x + 3)}\\

5^{(x - 2)} = (5- 3)^{x + 3}\\

5^{(x - 2)} = 5^{- 3x - 9}\\

x + 3x = - 9 + 2\\

4x = - 7\\

x = \frac{-7}{4} \)


61.

How many sides has a regular polygon whose interior angles are 120°each

.
A.

4

B.

5

C.

6

D.

3

Explanation.

Each exterior angle = 180 - 120=60

Exterior angle =360/n

60 = 360/n

n =360/6

= 6


62.
If duty is levied at 25%, find the duty to be added to a bill of N80.
.
A. N100
B. N20
C. N80
D. N150
Explanation.
25% of N80 = 25/100 \(\times\) 80

= N20.00

63.
Simplify 2.04 \(\times\) 3.7 (Leave your answer in 2 decimal place)
.
A. 7.51
B. 7.71
C. 5.5
D. 7.55
Explanation.
2.04 x 3.7
=
204
x 37
-----
7548 = 7.548

= 7.55(2dp)

64.

A public car dealer marked up the cost of a car at 30% in an attempt to make 20% gross profit. Due to the value of dollar, he now placed 20% discount on the car. What profit or loss will he make?

.
A.

3%

B.

2%

C.

4%

D.

1%

Explanation.

Let assume the cost price is 100%

Marked up price + cost price = 20 + 100 = 120%

Discount at 20% = 20/100 \(\times\) 120 % of cost price

Selling price = cost price - gain

= (120 - 24)% of cost price

= 96% of cost price

Loss = (100- 96)% of cost price

= 4% of cost price

- He will wake 4% loss


65.

Rationalize 5 ÷ (2 - \(\sqrt{3}\))

.
A.

3(2 + \(\sqrt{5}\))

B.

2(3 + \(\sqrt{5}\))

C.

5(2 + \(\sqrt{3}\))

D.

\(3\sqrt{3}\) + 1

Explanation.

5÷ (2 - \(\sqrt{3}\))

Using conjugate surds

[5 ÷ (2 - \(\sqrt{3}\))]\(\times\)(2 + \(\sqrt{3}\)) ÷ (2 + \(\sqrt{3}\))]

[5(2+\(\sqrt{3}\))÷((2- \(\sqrt{3}\)))2]

[5(2 +\(\sqrt{3}\)) ÷ (2)2- (\(\sqrt{3}\))2]

[5(2 +\(\sqrt{3}\)) ÷(4- 3)]

= 5(2 + \(\sqrt{3}\))


66.

Factorize \( a^2 - b^2 - 4a + 4 \)

.
A.

(a + b)(a - b)

B.

(a - 2 + b)

C.

(a + 1)(a - 2 + b)

D.

(a + b) 2

Explanation.

The trinomial = \( a^2 - 4a + 4 \\

a^2 - b^2- 4a + 4 = (a^2 - 4a + 4) - b^2 \\

(a^2 - 2a - 2a + 4) - b^2 \\

[a(a - 2)- 2(a - 2)] - b^2 \\

(a - 2)2 - b^2 \\

(a - 2 + b)(a - 2 - b )\)


67.
Given that tan x = \(\frac{2}{3}\), where 0o d" x d" 90o, Find the value of 2sinx.
.
A. \(\frac{2\sqrt{13}}{13}\)
B. \(\frac{3\sqrt{13}}{13}\)
C. \(\frac{4\sqrt{13}}{13}\)
D. \(\frac{6\sqrt{13}}{13}\)
Explanation.
tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\)

thus m2 = 22 + 32

= 4 + 9 = 13

m = \(\sqrt{13}\)

Hence, 2sin x = 2 x \(\frac{2}{m}\)

2 x\(\frac{2}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)

= \(\frac{4\sqrt{13}}{13}\)

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